Orientation of Surfaces I'm having a little trouble understanding how to orient a surface in $\mathbb{R}^3$ For example, how would I orient the ellipsoid given by:
$$x^2+y^2+z^2+xy+xz+yz=\frac12$$ 
for $(x,y,z) \in \mathbb{R}^3$ ?
 A: In elementary differential geometry of surfaces in $\Bbb R^3$ orienting a surface $S$ means defining a unit normal vector $n(p)$ at each point $p\in S$ such that $n$ variies continuously with $p$. The surface $S$ is called orientable if such an assignement is possible.
In your case $S$ is presented as level surface of the function
$$f(x,y,z):=x^2+y^2+z^2+xy+yz+zx\ .$$
In this case the vector $\nabla f(p)$ is normal to the tangent plane at $S$ and can be used for our purposes if it is $\ne0$. In order to check whether this condition is fulfilled we compute
$$\nabla f(x,y,z)=( 2x+y+z,\ x+2y+z,\ x+y+2z)$$
and easily check that the only point in $\Bbb R^3$ where $\nabla f$ vanishes is the origin $O$, and this is not a point of $S$. Therefore the unit vector
$$n(p):={\nabla f(p)\over|\nabla f(p)|}\qquad(p\in S)$$
provides an orientation of your ellipsoid $S$. 
The chosen orientation points outward. If for some reason you want to work with the opposite orientation you have to replace the above $n(p)$ by $-n(p)$.
A: As the OP put in a tag for differential forms, I suspect this final ingredient is desired: from Guillemin and Pollack Differential Topology pages 173-174, exercises 13 and 14. In particular 14 has 
$$  \omega = f_1 dy \wedge dz + f_2 dz \wedge dx +f_3 dx \wedge dy, $$ with a triple
$$ F = (f_1, f_2, f_3).  $$
Then the restriction of $\omega$ to the given surface $S$ is
$$F \cdot N \; dA$$
where $S$ is oriented and $N$ is the (outward) unit normal, as in Christian's answer. This should feel familiar, as it is exactly this setup in Stokes' Theorem for integrating a vector field over a surface.   
As a technical note, as long as the surface is smooth and strictly star-shaped around the origin (the radial vector is never tangent to the surface, never orthogonal to the normal), one may take
$$  \omega = x dy \wedge dz + y dz \wedge dx + z dx \wedge dy, $$
as a 2-form that never vanishes when restricted to the surface. 
A: The general form of an ellipsoid is:
$$({\bf x-v})^T A^{-1} ({\bf x-v})=1$$
From the form you presented, we can write the matrix $A^{-1}$ for your example as: 
$$A^{-1}= \left( \begin{array}{ccc}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2 \end{array} \right)$$
And you may find $A$:
$$A = \frac{1}{4}\left(
\begin{array}{ccc}
 3 & -1 & -1 \\
 -1 & 3 & -1 \\
 -1 & -1 & 3 \\
\end{array}
\right)$$
The eigenvectors of $A$ give you the axes of the ellipsoid:
$$(-1, 0, 1), (-1, 1, 0), (1, 1, 1)$$
And the eigenvalues $(1,1,1/4)$ give you the squares of the radii.
A: A simple way to do your integral is to use Stokes' theorem: the integral of $\omega$ over the surface of the ellipsoid is equal to the integral of $d\omega$ over the enclosed volume. Since you were given that $d\omega=dx\wedge dy\wedge dz$, which is the standard volume form in $R^3$, the integral is simply plus or minus the volume enclosed in the ellipsoid. To choose the sign, you must choose the orientation. It is ``standard" to choose the orientation determined by $dx\wedge dy\wedge dz$ in $R^3$. If you do this and if you then choose the induced orientation on the ellipsoid using the outward direction, the sign is positive. I wouldn't be surprised if the problem posed to you neglected to specify explicitly which orientation it wanted you to use.
